Complex Number

Problem 181: Show that

(a) Re $(iz)$ = -Im $z$

(b) Im $(iz)$ = Re $z$

where $z = x + iy$ .

Solution: It is important to note that $z$ is a complex number. Also, Re $z$ = $x$ and Im $z$ = $y$ . That is,

(a)

(1) $\begin{align*} iz & = i \cdot (x + iy) \\ & = ix + i^2 y \\ & = ix - y. \end{align*}$

Hence, Re $(iz)$ $= - y =$ -Im $z$ .

(b) Similarly, from above, Im $(iz)$ $= x =$ Re $z$ .